Factor the following expression: $5$ $x^2+$ $21$ $x+$ $4$
This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(4)} &=& 20 \\ {a} + {b} &=& & & {21} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $20$ and add them together. The factors that add up to ${21}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${20}$ $ \begin{eqnarray} {ab} &=& ({1})({20}) &=& 20 \\ {a} + {b} &=& {1} + {20} &=& 21 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {5}x^2 +{1}x +{20}x +{4} $ Group the terms so that there is a common factor in each group: $ ({5}x^2 +{1}x) + ({20}x +{4}) $ Factor out the common factors: $ x(5x + 1) + 4(5x + 1) $ Notice how $(5x + 1)$ has become a common factor. Factor this out to find the answer. $(5x + 1)(x + 4)$